Question

demonstrati ca x,y si z sunt numere reale pozitive,astfel incat x+y+z=1344, atunci$\sqrt{x(y+z)}$+$\sqrt{y(x+z)}$+$\sqrt{z(x+y)}$≤2016.

Answer 1

Cmmdc(1344;2016)=?
1344=(2^6)*3*7
2016=(2^5)*(3^2)*7, unde "^"=la puterea, "*"=inmultit, "V"=radical
Cmmdc(1344;2016)=(2^5)*3*7=672

1344=672*2
2016=672*3=[(672*2)/2]*3=(1344/2)*3=(3/2)*1344=
=(3/2)*(x+y+z)

In inecuatie inlocuim 2016=(3/2)*(x+y+z)

V[x(y+z)]+V[y(x+z)]+V[z(x+y)]<=(3/2)*(x+y+z) l *2
2V[x(y+z)]+2V[y(x+z)]+2V[z(x+y)]<=3(x+y+z)
3(x+y+z)-2V[x(y+z)]-2V[y(x+z)]-2V[z(x+y)]>=0
Pe 3x+3y+3z il impartim convenabil si grupam in paranteze:
{x-2V[x(y+z)]+y+z}+{y-2V[y(x+z)]+x+z}+{z-2V[z(x+y)]+x+y}>=0
Se observa ca parantezele sunt patrate:
[Vx-V(y+z)]^2 +[Vy-V(x+z)]^2 +[Vz-V(x+y)]^2>=0
x, y, z sunt reale pozitive, deci radicalii au sens.
Suma a 3 patrate este intotdeauna un nr. pozitiv.
Concluzie: inecuatia data, conditionata de egalitatea din enunt, este adevarata pt. orice x,y,z reale, pozitive.